Find #a, b# so that the system #x+y+z=6#, #x+2y+3z=10#, #x+2y+az=b# has an unique solution?

2 Answers
Eddie
Apr 21, 2017

You have:

#((1,1,1),(1,2,3),(1,2,a)) mathbf x = ((6),(10),(b))#

Now if #a = 3#, you have problems because the 2nd and 3rd columns vectors are linearly dependent , as are the 2nd and 3rd row vectors.

Ouside of that, you are good to go :)

A cool way to go, if in doubt, is to row-reduce the augmented matrix:

So:

#((1,1,1),(1,2,3),(1,2,a)) ((6),(10),(b))#

#R2 to R2 - R1; R3 to R3 - R1#

#((1,1,1),(0,1,2),(0,1,a-1)) ((6),(4),(b-6))#

# R3 to R3 - R2#

#((1,1,1),(0,1,2),(0,0,a-3)) ((6),(4),(b-10))#

It is only if #a = 3# that we do not have 3 pivots and therefore a solution.

By back substitution, if #a ne 3#, then: #z = (b-10)/(a-3)#. IOW, we don't really care what #b# amounts to :)

Raj
Apr 22, 2017

Please check below

Explanation:

I tried by using Cramer's Rule with following results:

#|D| = |A| = [(1, 1, 1),(1, 2, 3),(1, 2, a)]##= a- 3, #which is# != 0 iff a != 3#

Therefore we can find the other determinants:

#|D_1| = [(6, 1, 1),(10, 2, 3),(b, 2, a)]##= 12a+b-36#

#|D_2| = [(1, 6, 1),(1, 10, 3),(1, b, a)]##= 4a-2b+8#

#|D_3| = [(1, 1, 6),(1, 2, 10),(1, 2, b)]##= b#

Now,

#x = D_1/D = (12a+b-36)/(a-3)#,
#y = D_2/D = (4a-2b+8)/(a-3)#
#z = D_3/D = b/(a-3)#

What should be the statement to determine the value of #a, b# which makes these values unique for the given set?

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